Packet Substitution Solutions - Integration by Substitution Solutions Note in(1 through(6 we nd the answer by setting du equal to the derivative of the

# Packet Substitution Solutions - Integration by Substitution...

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Integration by Substitution Solutions Note: in (1) through (6), we find the answer by setting du equal to the derivative of the function, multiplied by dx . 1. u = 4 x + 1 du = 4 dx 2. u = x 4 + 2 du = 4 x 3 dx 3. u = 3 x 2 + 1 du = 6 xdx 4. u = ln( x ) du = 1 x dx 5. u = 1 /x = x - 1 du = - x - 2 dx 6. u = e x + 2 du = e x dx 7. u = 4 x + 1 du = 4 dx du/ 4 = dx R (4 x + 1) 4 dx = R u 4 du 4 = 1 4 R u 4 du = 1 4 · u 5 5 + C = 1 20 (4 x + 1) 5 + C 8. u = - 2 x + 3 du = - 2 dx - du/ 2 = dx R e - 2 x +3 dx = R e u · - du 2 = - 1 2 R e u du = - 1 2 e u + C = - 1 2 e - 2 x +3 + C 9. Note: in problems like this one, it may not be immediately obvious what to set u equal to (e.g., here, you might set u = 5 x 3 or u = x 4 +2). You can try trial and error, by setting u equal to something and trying to solve 1
2 the problem, and then starting over if you can’t cancel out all of the ‘ x ’s; another tip is to look for a function whose derivative you see in the integral. So if you set u = x 4 + 2, the derivative is 4 x 3 , which is similar to the 5 x 3 we already see in the integral; this is a good indicator that u = x 4 + 2 is the right choice.